Answer:
1.72x10⁻⁵ g
Explanation:
To solve this problem we use the PV=nRT equation, where:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 25 °C ⇒ (25+273.16) = 298.16 K
And we <u>solve for n</u>:
- 1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
Finally we <u>convert moles of helium to grams</u>, using its <em>molar mass</em>:
- 4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g
I'm guessing water vapour. i can't think of any other gas limited to only hydrogen and oxygen
hope it helps :)
Can you use a calculator? If so, demos works great for problems like these!
Basis: 1 L of the substance.
(1.202 g/mL) x (1000 mL) = 1202 g
mass solute = (1202 g) x 0.2 = 240.2 g
mass solvent = 1202 g x 0.8 = 961.6 g
moles KI = (240.2 g) x (1 mole / 166 g) = 1.45 moles
moles water = (961.6 g) x (1 mole / 18 g) = 53.42 moles
1. Molality = moles solute / kg solvent
= 1.45 moles / 0.9616 kg = 1.5 m
2. Molarity = moles solute / L solution
= 1.45 moles / 1 L solution = 1.45 M
3. molar mass = mole solute / total moles
= 1.45 moles / (1.45 moles + 53.42 moles) = 0.0264
A grey coloured rock with amphibole and intermediate plagioclase like an andesine would classify as an intermediate rock by Bowen's Reaction Series and by the classification of igneous rocks would probably be like a diorite which is intermediate between a gabbro and a granite. A diorite essentially has no quartz but has the silicates amphibole (like hornblende), mica perhaps a little pyroxene and andesine plagioclase.
